1986 CHAPTER 61. PROBABILITY IN INFINITE DIMENSIONS

for all A ∈ G . The letting A≡ Z−1(

B(

an,||an||

4

))it follows

0 =∫

AZ−an +andP

and so

||an||P(A) =∣∣∣∣∣∣∣∣∫A

andP∣∣∣∣∣∣∣∣= ∣∣∣∣∣∣∣∣∫A

(an−Z)dP∣∣∣∣∣∣∣∣

≤∫

Z−1(

B(

an,||an ||

4

)) ||an−Z||dP≤ ||an||4

P(A)

which is a contradiction unless P(A) = 0. Therefore, letting

N ≡ ∪∞n=1Z−1

(B(

an,||an||

4

))= Z−1 (E \{0})

it follows N has measure zero and so Z = 0 a.e. This proves uniqueness because if Z,Z′

both hold, then from the above argument, Z−Z′ = 0 a.e.Next I will show Z exists. To do this recall Theorem 21.2.4 on Page 652 which is stated

below for convenience.

Theorem 61.1.2 An E valued function, X, is Bochner integrable if and only if X is stronglymeasurable and ∫

Ω

||X (ω)||dP < ∞. (61.1.1)

In this case there exists a sequence of simple functions {Xn} satisfying∫Ω

||Xn (ω)−Xm (ω)||dP→ 0 as m,n→ ∞. (61.1.2)

Xn (ω) converging pointwise to X (ω),

||Xn (ω)|| ≤ 2 ||X (ω)|| (61.1.3)

andlimn→∞

∫Ω

||X (ω)−Xn (ω)||dP = 0. (61.1.4)

Now let {Xn} be the simple functions just defined and let

Xn (ω) =m

∑k=1

xkXFk (ω)

where Fk ∈F , the Fk being disjoint. Then define

Zn ≡m

∑k=1

xkE(XFk |G

).